[MUSIC] The question which we would like to solve now How to construct cusp product Theta a1. So, theta a8. Cusp being the correspondent, Jacobi modular form is a cusp form. Proposition Let a and b be positive. Then the product of two thetas, theta a x theta b. The Jacobi form of weight 1 index a to the square plus b to the square over 2 with respect to the character not multiply system (v eta to the power 6 x v H to the power a + b). This product is a cusp If and only if the number a times b over the square of the greatest common divisor a, b is even. The proof is rather simple. Let us calculate Ze for the expansion of this product. With a respect to the 2 in this instance. The connecting symbol is a character. Q to the power n1 to the square + n2 to the square / 8, r to the power. an1 + bn2 / 2. As usual, r = e to the power 2 pi i z, and q is e to the power 2 pi i tau. So this is our 4e and k efficient. Certainly we can make a re-summation to fix the same N/2 and the same, l/2. Then the hyperbolic norm. Over the index for Norm 0 Fourier coefficient. Sorry here not n over 2 but This is n over 8 the node of the corresponding index, and the index here, is the first part. The 5th called into this index n over 2 in the second l over 2, this is the index of non-0 of Fourier coefficient Is equal, 4 times the index of Jacobi form * N /8. L /2, to the square. This our usual hyperbolic norm for nm minus n2 square. In this particular case we have 1/4 because 4/16 is 4 x N1 to the square plus N2 to the square x a square + b square- a plus 1 plus b squared. Let me correct this formula a little bit. And this is certainly equal. To 1/4(n1b- N2a) to the square. Please check this. It's very simple calculation. And certainly, this known is greater or equal to 0. And this inequality reflects the product of two holomorphic Jacobi forms. But now let's assume that a and b are. This gives us no restriction at all. Then, a x b is even. But, n1 and n2, they both, are odd. Because we have this character in the Fourier summation. So the Kronecker symbol is 0 if the product n1 x n2 is even. But now, a or b is even, and n1 and n2 are odd. So we see that this square is strictly positive. Due to these two It gives us the proof of the proposition. So if the product of two jacobi is a cusp form under this condition. And now we can formulate the correspondent result about the product of 8 or more generally kappa theta series corollary. I would like to use the following notation. That A is the vector, A 1 A, K. All A, E, are strictly positive. Then theta with index the vector a is by definition the product of k theta functions. Let's for simplicity, there is not really friction. We, assume that the greatest common divisors, maybe I write it explicit form. The greatest common divisor of a1, ak, is equal to 1. Then. Theta a Is a cusp Jacobi form If At least One ai is even. Moreover, this function, theta a, is a Jacobi form of weight K/2 of index A1 to the square AK to the square/2 with trivial character. If and only if K is congruous to 0, modular 8, then the sum aK is even and under this condition this Jacobi form is a cusp form. If and only if At least one a:e is even. So we solve the problem. We describe, all theta product. We check as form. And I can construct the simplest examples. We take, with a vector a, 6 units. And 2, 2. There's the correspondent. Data function. Eta to the power 6 x theta 2 to the square is jacobi form of J4 and this is 4 and index 6 + 4 + 4/2. And index 7. This is the first jacobi cusp form which we can construct using the theta product. [MUSIC]